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2 years ago

How many grams of each of the following substances will dissolve in 100 ml of cold water; Ce(IO3)4, RaSO4, (NH4)2SeO4?

Chemistry

2 answers:

DIA [1.3K]2 years ago

5 0

Answer: $Ce(IO_3)_4$ = 0.123g , $RaSO_4$ = $2.1*10^-^4g$ and $(NH_4)2SeO_4$ = 96 g

Explanations: The question asks about the solubility of each of the compounds in grams per 100 mL of water. density of water is 1 gram per mL. So, 100 mL of water would be same as 100 g water.

In solubility charts, the solubilities are also shown in grams of compound per 100 grams of water.

Solubility of $Ce(IO_3)_4$ is 0.123 g, solubility of $RaSO_4$ is 0.00021 g and the solubility of $(NH_4)2SeO_4$ is 96 g per 100 g of water.

Since the question also asks about grams of each soluble in 100 mL that is 100 g of water, the grams of each salts will be same as their above solubilities that is, $Ce(IO_3)_4$ is 0.123 g, $RaSO_4$ is 0.00021 g and the solubility of $(NH_4)2SeO_4$ is 96 g.

ipn [44]2 years ago

4 0

The solubility of Ce(IO3)4, RaSO4, (NH4)2SeO4 in water are 0.123, 2.1x10^-4 and 96 in units of g/100 mL. Therefore, given 100 mL of water 0.123 g of Ce(IO3)3, 2.1x10^-4 g of RaSO4 and 96 g of (NH4)2SeO4 can be dissolve. Hope this helps.

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Answer is: A. Chemical energy to electromagnetic energy and thermal energy.

Balanced chemical reaction: 2Mg(s) + O₂(g) → 2MgO(s) + energy.

This is chemical change (chemical reaction), because new substance (magnesium oxide MgO) is formed, the atoms are rearranged and the reaction is followed by an energy change (exothermic reaction because energy is released).

Chemical changes (chemical synthesis) is when a substance combines with another (in this example magnesium and oxygen) to form a new substance.

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2 years ago

Read 2 more answers

A student pours exactly 26.9 mL of HCl acid of unknown molarity into a beaker. The student then adds 2 drops of the indicator an

Assoli18 [71]

a.
Acids react with bases and give salt and water and the products.

Hence, HCl reacts with NaOH and gives NaCl salt and H₂O as the products. The reaction is,
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

To balance the reaction equation, both sides hould have same number of elements.

Left hand side, Right hand side,
H atoms = 2 H atoms = 2
Cl atoms = 1 Cl atoms = 1
Na atoms = 1 Na atoms = 1
O atoms = 1 O atoms = 1

Hence, the reaction equation is already balanced.

b.
Molarity (M)= moles of solute (mol) / Volume of the solution (L)

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Molarity of NaOH = 0.13 M
Volume of NaOH added = 43.7 mL
Hence, moles of NaOH added = 0.13 M x 43.7 x 10⁻³ L
= 5.681 x 10⁻³ mol

Stoichiometric ratio between NaOH and HCl is 1 : 1

Hence, moles of HCl = moles of NaOH
= 5.681 x 10⁻³ mol

5.681 x 10⁻³ mol of HCl was in 26.9 mL.

Hence, molarity of HCl = 5.681 x 10⁻³ mol / 26.9 x 10⁻³ L
= 0.21 M

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2 years ago

What types of compounds are CaCl2, Cu, C2H6, respectively.

mina [271]

Answer:

Ionic, metal, organic

Explanation:

In this case, we have to analyze each compound:

-) $CaCl_2$

In this compound, we have a non-metal atom (Cl) and a metal atom (Ca) . So, we will have a high electronegativity difference between these atoms, With this in mind, we will have an ionic bond. Ions can be produced:

$CaCl_2~->~Ca^+^2~+~2Cl^-$

The cation would be $Ca^+^2$ and the anion is $Cl^-$ . So, we will have an ionic compound.

-) $Cu$

In this case, we have a single atom. If we check the periodic table we will find this atom in the transition metals section (in the middle of the periodic table). So, this indicates that Cu (Copper) is a metal.

-) $C_2H_6$

In this molecule, we have single bonds between carbon and hydrogen. The electronegativity difference between C and H are not high enough to produce ions. So, with this in mind, we will have covalent bonds. This is the main characteristic of organic compounds.  (See figure 1)

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2 years ago

Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of

shutvik [7]

Answer:

79%

Explanation:

1) Balanced chemical equation:

2S + 3O₂ → 2SO₃

2) Mole ratio:

2 mol S : 3 mol O₂ : 2 mol SO₃

3) Limiting reactant:

Number of moles of O₂

n = 6.0 g / 32.0 g/mol = 0.1875 mol O₂

Number of moles of S:

n = 7.0 g / 32.065 g/mol = 0.2183 mol S

Ratios:

Actual ratio: 0.1875 mol O₂ / 0.2183 mol S =0.859

Theoretical ratio: 3 mol O₂ / 2 mol S = 1.5

Since there is a smaller proportion of O₂ (0.859) than the theoretical ratio (1.5), O₂ will be used before all S be consumed, and O₂ is the limiting reactant.

4) Calcuate theoretical yield (using the limiting reactant):

0.1875 mol O₂ / x = 3 mol O₂ / 2 mol SO₃

x = 0.1875 × 2 / 3 mol SO₃ = 0.125 mol SO₃

5) Yield in grams:

mass = number of moles × molar mass = 0.125 mol × 80.06 g/mol = 10.0 g

6) Percent yield:

Percent yield, % = (actual yield / theoretical yield) × 100

% = (7.9 g / 10.0 g) × 100 = 79%

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2 years ago

For a school event 1/6 of the athletic field is reversed for the fifth -grade classes the reserved part of the field is divided

SVETLANKA909090 [29]

Answer:

$\frac{1}{24}$

Explanation:

Given:

For a school event, 1/6 of the athletic field is reserved for the fifth -grade classes and the reserved part of the field is divided equally among the 4 fifth grade classes in the school.

To find: fraction of the whole athletic field reserved for each fifth class

Solution:

Fraction of the whole athletic field reserved for four fifth classes = $\frac{1}{6}$

So, fraction of the whole athletic field reserved for each fifth class = $\frac{1}{4}(\frac{1}{6})=\frac{1}{24}$

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2 years ago